Answer: \(v=2,0,2.5,-3.75\) on the four intervals; total distance \(300\) m; acceleration \(-27\text{ m s}^{-2}\).

(a)(i) Velocity is the gradient of the displacement-time graph.

From \(0\leq t\leq50\),

\(v=\frac{100-0}{50-0}=2.\)

From \(50\leq t\leq65\), the displacement is constant, so

\(v=0.\)

From \(65\leq t\leq85\),

\(v=\frac{150-100}{85-65}=2.5.\)

From \(85\leq t\leq125\),

\(v=\frac{0-150}{125-85}=-3.75.\)

So the \(v\)-\(t\) graph consists of horizontal line segments at \(2\), \(0\), \(2.5\) and \(-3.75\) on these four intervals.

(a)(ii) The total distance is the total change in displacement, ignoring direction:

\(100+0+50+150=300.\)

So the total distance is \(300\) m.

(b) Given

\(x=6\cos\left(3t+\frac{\pi}{3}\right),\)

differentiate once:

\(\frac{dx}{dt}=-18\sin\left(3t+\frac{\pi}{3}\right).\)

Differentiate again:

\(\frac{d^2x}{dt^2}=-54\cos\left(3t+\frac{\pi}{3}\right).\)

When \(t=\frac{2\pi}{3}\),

\(3t+\frac{\pi}{3}=2\pi+\frac{\pi}{3}=\frac{7\pi}{3}.\)

Since \(\cos\frac{7\pi}{3}=\frac12\),

\(\frac{d^2x}{dt^2}=-54\left(\frac12\right)=-27.\)